Subject

Chemistry

Class

NEET Class 12

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

 Multiple Choice QuestionsMultiple Choice Questions

1.

A buffer solution is prepared in which the concentration of NH3 is 0.30 M and the concentration of NH4 is 0.20 M. If the equilibrium constant, Kb for NH3 equals 1.8 x 10-5, what is the pH of this solution?

log ( 2.7 = 0.43)

  • 9.43

  • 11.72

  • 8.73

  • 8.73


A.

9.43

pOH space equals space pK straight b space plus space log space fraction numerator left square bracket Salt right square bracket over denominator left square bracket Base right square bracket end fraction

equals negative space log space straight K subscript straight b space plus space log space fraction numerator open square brackets Salt close square brackets over denominator left square bracket Base right square bracket end fraction

equals negative space log space 1.8 space straight x space 10 to the power of negative 5 end exponent space plus space log space fraction numerator 0.20 over denominator 0.30 end fraction

space equals space minus 5 minus 0.25 space plus space left parenthesis negative 0.176 space right parenthesis

therefore comma

4.75 space minus space 0.176 space equals space 4.57

therefore comma space pH space equals space 14 space minus space 4.57 space equals space 9.43
981 Views

2.

Enthalpy change for the reaction,

4H (g) → 2H2 (g) is - 869.6 kJ

The dissociation energy of H - H bond is

  • -869.6 kJ

  • +434.8 kJ

  • +217.4 kJ

  • +217.4 kJ


B.

+434.8 kJ

4H (g) → 2H2 (g); ΔH = - 869.6 kJ

2H2 (g) → 4 H (g); ΔH =  869.6 kJ

H2 (g) → 2 H (g); ΔH = 869.6/2 = 434.8 kJ

774 Views

3.

Which of the two ions from the list given below have the geometry that is explained by the same hybridization of orbitals,
NO subscript 2 superscript minus comma space NO subscript 3 superscript minus comma space NH subscript 2 superscript minus comma space NH subscript 4 superscript plus comma space SCN to the power of minus ?

  • NH subscript 4 superscript plus space and space NO subscript 3 superscript minus
  • SCN to the power of minus space and space NH subscript 2 superscript minus
  • NO subscript 2 superscript minus space and space NH subscript 2 superscript minus
  • NO subscript 2 superscript minus space and space NH subscript 2 superscript minus


D.

NO subscript 2 superscript minus space and space NH subscript 2 superscript minus

Hybridization of the given molecule is 

NO subscript 2 superscript minus space rightwards arrow space sp squared
NO subscript 3 superscript minus space rightwards arrow space sp squared
NH subscript 2 superscript plus space rightwards arrow space sp cubed
NH subscript 4 superscript minus space rightwards arrow space sp cubed
SCN to the power of plus space rightwards arrow sp

therefore, NO subscript 2 superscript minus space and space NO subscript 3 superscript minus both have the same hybridization.

816 Views

4.

The Correct order of increasing bond length of C - H, C-O, C - C and C = C is


  • C - C < C=C < C - O < C - H

  • C - O < C - H < C - C < C = C

  • C - H < C - O <  C - C < C= C

  • C - H < C - O <  C - C < C= C


D.

C - H < C - O <  C - C < C= C

C - H: 0.109 nm

C = C : 0.134 nm

C - O: 0.143 nm

C - C : 0.154 nm

Therefore, Bond length order is 

C - H < C = C < C- O < C - C 

546 Views

5.

A gaseous mixture was prepared by taking equal moles of CO and N2, If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen (N2) in the mixture is

  • 0.8 atm

  • 0.9 atm

  • 1 atm

  • 1 atm


C.

1 atm

because space CO presuperscript straight n space equals space straight N presuperscript straight n subscript 2
therefore space straight p subscript CO space equals space straight p subscript straight N subscript 2 end subscript

Given comma space straight p subscript CO space plus space straight p subscript straight N subscript 2 space end subscript equals space 1 space atm

Or space space 2 straight p subscript straight N subscript 2 end subscript space equals space 1 space atm

straight p subscript straight N subscript 2 end subscript space equals space 0.5 space atm
1793 Views

6.

Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively. The molecular mass of A is 49u. Molecular mass of B will be

  • 12.25 u

  • 6.50 u 

  • 25.00 u

  • 25.00 u


A.

12.25 u

fraction numerator Rate space of space diffusion space of space straight I space gas over denominator Rate space of space diffusion space II space gas end fraction space equals space square root of M subscript B over M subscript A end root
fraction numerator begin display style bevelled v subscript 1 over t subscript 1 end style over denominator begin display style bevelled v subscript 2 over t subscript 2 end style end fraction space equals space space square root of M subscript B over 49 end root
since space the space volumes space are space equal space straight V subscript 1 space equals space straight V subscript 2
fraction numerator begin display style 10 end style over denominator begin display style 20 end style end fraction space equals space space square root of M subscript B over 49 end root

fraction numerator begin display style 10 end style over denominator begin display style 20 end style end fraction space equals space space fraction numerator square root of straight M subscript 2 end root over denominator 7 end fraction

square root of straight M subscript 2 end root space equals fraction numerator 10 space straight x space 7 over denominator 20 end fraction space equals space 3.5

square root of straight M subscript 2 end root space equals space 3.5
Thus comma
straight M subscript 2 space equals space 12.25 space straight u
445 Views

7.

If the enthalpy change for the transition of liquid water to steam is 30kJ mol-1 at 27oC, the entropy change for the process would be. 

  • 1.0 J mol- K-1

  • 0.1 J mol-1 K-1

  • 100 J  mol-1 K-1

  • 100 J  mol-1 K-1


C.

100 J  mol-1 K-1

increment straight G to the power of straight o space equals space increment straight H to the power of straight o space minus space straight T increment straight S to the power of straight o

Given comma space increment straight H subscript vap space equals space 30 space kJ space mol to the power of negative 1 end exponent

increment straight G to the power of straight o space equals space 0 space at space equilibrium comma

increment straight S subscript vap space equals space fraction numerator increment straight H subscript vap over denominator straight T end fraction

space equals space fraction numerator 30 space straight x space 10 cubed space straight x space straight J space mol to the power of negative 1 end exponent over denominator 300 space straight K end fraction
space equals space 100 space straight J space mol to the power of negative 1 end exponent space straight K to the power of negative 1 end exponent
553 Views

8.

If x is the amount of adsorbate and m is the amount of adsorbent, which of the following relations is not related to adsorption process?

  • straight x over straight m space equals space straight f left parenthesis straight T right parenthesis space at space constant space straight p
  • p = f(T) at constant (x/m)

  • straight x over straight m space proportional to fraction numerator space straight p over denominator straight T end fraction
  • straight x over straight m space proportional to fraction numerator space straight p over denominator straight T end fraction

C.

straight x over straight m space proportional to fraction numerator space straight p over denominator straight T end fraction

The correct relation is straight x over straight m space proportional to fraction numerator space straight p over denominator straight T end fraction

462 Views

9.

Which of the following is least likely to behave as Lewis base? 

  • NH3

  • BF3

  • OH-

  • OH-


B.

BF3

BF3 is an electron deficient species, thus behaves like a Lewis acid.

475 Views

10.

Which of the following is the correct option for free expansion of an ideal gas under an adiabatic condition?

  • straight q space not equal to space 0 comma space increment straight T space equals space 0 comma space straight W space equals 0
  • straight q space equals space 0 comma space increment straight T space equals space 0 comma space straight W space equals space 0
  • straight q space equals 0 comma space increment straight T space less than thin space 0 comma space straight W not equal to 0
  • straight q space equals 0 comma space increment straight T space less than thin space 0 comma space straight W not equal to 0

B.

straight q space equals space 0 comma space increment straight T space equals space 0 comma space straight W space equals space 0

For an adiabatic process, q = 0 and for free expansion, W = 0,

Therefore ΔT = 0.

750 Views